算法第八周解题报告

98. Validate Binary Search Tree

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than the node's key.
• Both the left and right subtrees must also be binary search trees.

Example 1:

    2
   / \
  1   3

Binary tree [2,1,3], return true.

Example 2:

    1
   / \
  2   3

Binary tree [1,2,3], return false.

解题思路：本人的思路是通过后序遍历，对于每个非叶子结点:

1.判断他们的左右子树是否是搜索二叉树

2.取回左右子树的最大最小值，判断以这个节点为根的树是否是搜索二叉树。

算法代码如下:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isValidBST(TreeNode* root) {
        int max, min;
        return travel(root, min, max);
    }
    bool travel(TreeNode* root, int & min, int & max){
        if (root == NULL) {
            max = INT_MIN;
            min = INT_MAX;
            return true;
        }
        int leftMin, leftMax, rMin, rMax;
        bool isBSTree = travel(root -> left, leftMin, leftMax);
        if(isBSTree == false) return false;
        isBSTree = travel(root -> right, rMin, rMax);
        if(isBSTree == false)return false;
        if(root -> left != NULL && root -> val <= leftMax) return false;
        if(root -> right != NULL && root -> val >= rMin) return false;
        max = leftMax > rMax ? leftMax : rMax;
        max = root -> val > max ? root -> val : max;
        min = leftMin < rMin ? leftMin : rMin;
        min = root -> val < min ? root -> val : min;
        return true;
    }
};

结果如下: